Question

1) When quantities cannot be formulated without identifying their directions, they are called as- 2)A straight line graph has constant - 3)Graph for body thrown upwards will be a linear graph (true or false)- 4)Slope of a velocity time graph is used to determine- 5)When speed changes from 10m/s to 30m/s in 10 seconds, acceleration is - 6)Distance of 50 meters is covered by a truck after starting from rest in 10 seconds. Acceleration of truck will be- 7)A toy train reaches to a speed of 4 kmph from 1 kmph in 5 seconds. Calculate Rate of change in speed -

Answer 1

(i) When quantities cannot be formulated without identifying their directions, they are called as Vector Quantities

(vi) The acceleration of the truck will be 1 m/s²

(vii) Rate of change of speed is 0.167 m/s²

(ii) A straight line graph has constant slope

(iii) False   (Graph of a body thrown upwards will not be a linear graph)

(iv) Slope of a velocity time graph is used to determine acceleration

(v) When speed changes from 10 m/s to 30 m/s in 10 seconds, the acceleration is  2 m/s²

(vi) Acceleration of the truck will be 1 m/s²

(vii) Rate of change of speed = 0.167 m/s²

Explanation:

(1) Vector quantities are quantities that have both magnitude and direction

(2) A straight line is a line that has constant slope

(3) A body thrown upwards with initial velocity u, going upto height h in time t can be written as

$h=ut-\frac{1}{2}gt^2$

Clearly it can be seen that the graph between the height h and time t is not a straight line rather a parabola because here the degree of the equation is 2

(4) We know that acceleration is given by

$a=\frac{dv}{dt}$

mathematically dv/dt is slope of the curve between velocity v and time t

Thus, slope of velocity time graph is used to determine acceleration

(5) Change in speed

$\Delta v=30-10=20$ m/s

time $\Delta t=10$ seconds

Acceleration

$a=\frac{\Delta v}{\Delta t}$

$\implies a=\frac{20}{10}$

$\implies a=2$ m/s²

(6) Given

$s=50$ m

$u=0$

$t=10$ seconds

Using

$s=ut+\frac{1}{2}at^2$

$50=0\times 10+\frac{1}{2}a\times 10^2$

$\implies 100=100a$

$\implies a=1$ m/s²

(7) Change in speed

$\Delta v=4-1=3$ kmph

$\implies \Delta v=3\times (5/18)$ m/s

time $\Delta t=5$ seconds

Rate of change of speed = acceleration

$=\frac{\Delta v}{\Delta t}$

$=\frac{15/18}{5}$

$=0.167$ m/s²

Hope this answer is helpful.

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Answer 2

Answer:

Heya mate!!!

Explanation:

So here is your answer :

_______________________

(i) When quantities cannot be formulated without identifying their directions, they are called as Vector Quantities

(ii) A straight line graph has constant slope

(iii) False

(iv) Slope of a velocity time graph is used to determine acceleration

(v) When speed changes from 10 ms⁻¹ to 30 ms⁻¹ in 10 seconds, the acceleration is  2 ms⁻²

(vi) The acceleration will be 1 ms⁻²

(vii) Rate of change of speed is 0.167 ms⁻²

________________________

Hope it helps yaa mate ♡♡♡ !!

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